For the following sine waveform, if E = 7.9 V, what is the voltage at t = 3.8ms? (write your answer to two decimal places)

Break Down the Problem

1. Identify the sine wave equation: $V(t) = E \cdot \sin(\omega t + \phi)$, where:

   • $E$ is the peak voltage,
   • $\omega$ is the angular frequency,
   • $\phi$ is the phase angle,
   • $t$ is time.

2. Given:

   • Peak voltage, $E = 7.9 \, \text{V}$.
   • Time, $t = 3.8 \, \text{ms} = 0.0038 \, \text{s}$.

3. Assume the sine wave has a fundamental frequency of $f$. In a typical case, $\omega = 2\pi f$ (but we will need the frequency to compute the voltage accurately).

Relevant Concepts

1. Assume a standard frequency; let's say the sine wave has a frequency of $f = 50 \, \text{Hz}$ for this example. Therefore,
   • $\omega = 2\pi f = 2\pi (50) = 100\pi \, \text{rad/s}$.

Analysis and Detail

1. Plug the values into the sine wave equation:
   • $V(t) = 7.9 \cdot \sin(100\pi \cdot 0.0038)$.
   • Calculate $100\pi \cdot 0.0038$: $100\pi \cdot 0.0038 \approx 1.192 \, \text{rad}$
   • Now compute the sine: $\sin(1.192) \approx 0.924$
   • Therefore, calculate $V(t)$: $V(t) = 7.9 \cdot 0.924 \approx 7.2926 \, \text{V}$

Verify and Summarize

1. Rounding the result to two decimal places: $V(t) \approx 7.29 \, \text{V}$

Final Answer

The voltage at $t = 3.8 \, \text{ms}$ is approximately 7.29 V.